Japanese - Math Olympiad Algebra Problem

[@ungarlinski7965]

Rational roots theorem to see that -1 is root. Synthetic division .... Child's play.

[@johnstanley5692]

use x^6 = 1+(x-1)*(1+x+x^2+x^3+x^4+x^5). Given (1+x+..+x^5) = 0 (x~=0) gives x^6 = 1 = exp(i*2*n*pi) . Hence x=exp(i*n*pi/3) where n= 1:5. This method gives same result but seems easier to use.

[@hoffiehoffmann4286]

-1

[@TomJones-tx7pb]

That is how I did it to generate all of the roots of the equation.

[@jeanvandevelde7712]

Much easier to multiply by x-1 to get x^6-1 =0 and x not equal to 1.

[@Tengrikut121]

I directly put -1 and got the correct result without any calculation

[@Whyrwehere275]

3 odd exponents gave it away.

[@aines7958]

Excellent 👍👍👍👍❤🌹🇩🇿

[@elephantsarenuts5161]

"As is obvious to the casual observer ..." was the phrase we used in my algebra class ... 50 years ago.

[@user-xu9qc9tc9q]

共通因数でくくる方法もあるけど、右辺を左辺に移項させて、両辺に(x-1)をかけると、x^6-1＝0となり、x＝1以外の複素数解になりますね。 ここで複素平面上に半径1の円を書いて、ル·モアブルの定理より、中心角がπ/6の正6角形を書けば、自ずと分かりますね。

[@user-vd9so1wl3u]

x^4 + x^2 + 1 is always greater then 0, so, only x+1 can be equel to 0. So, x = -1.

[@maxkim2479]

It is a summ of geometric progression.

[@user-dq3uh6ee5w]

-1 (real value).

[@skearns23]

Or, Y = x^4 + x^3 + x^2 + x + 1 so xy + 1 = 0 and x^5 + y = 0 sub y = -x^5 into xy + 1 = 0 get x^6 = 1

[@luonglethanh9492]

X=-1

[@mjpottertx]

Easy enough by inspection.

[@robertloveless4938]

Knew this IMMEDIATELY.